Prove that any integer $n$ whose prime factorization is $n=p_1^{a_1}p_2^{a_2}…p_k^{a_k}$ is a perfect $m$th...
$begingroup$
I am a high school student self-studying number theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics:
Prove that any integer $n$ whose prime factorization is $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ is a perfect square iff $2|a_i$, and in general $n$ is an $m^{th}$ power iff all $a_i$ is a multiple of $m$.
I managed to prove this for $m=2$, as the question said, but am unable to even get started on the second part, because my proof required an assumption which I do not know if holds for the general case also.
Please check the proof below:
$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ where $n$ is a perfect square.
Let us assume, if possble, that there exist some powers $a_igeq 0$ such that $a_i=2b_i+1$.
We know that the number of factors $tau(n)=(a_1+1)(a_2+1)...(a_k+1)$. We can write each $a_i$ as $a_i= 2b_i$ or $a_i=2b_i+1$.
Now $tau(n)=(2b_1+1)(2b_2+1)...(2b_i+1+1)...(2b_k+1)$.
$Rightarrow 2|tau (n)$
$color{red}{textrm{However we know that every perfect square has an odd number of factors.}}$ This contradicts our assumption that $n$ is a perfect square.
Thus, our assumption is wrong and no such $a_i=2b_i+1$ exists.
QED.
Can the assumption highlighted in red be generalized for any number which is an $m^{th}$ power? And if not, is there a better way of doing it than my proof?
elementary-number-theory proof-verification
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
$endgroup$
|
show 1 more comment
$begingroup$
I am a high school student self-studying number theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics:
Prove that any integer $n$ whose prime factorization is $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ is a perfect square iff $2|a_i$, and in general $n$ is an $m^{th}$ power iff all $a_i$ is a multiple of $m$.
I managed to prove this for $m=2$, as the question said, but am unable to even get started on the second part, because my proof required an assumption which I do not know if holds for the general case also.
Please check the proof below:
$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ where $n$ is a perfect square.
Let us assume, if possble, that there exist some powers $a_igeq 0$ such that $a_i=2b_i+1$.
We know that the number of factors $tau(n)=(a_1+1)(a_2+1)...(a_k+1)$. We can write each $a_i$ as $a_i= 2b_i$ or $a_i=2b_i+1$.
Now $tau(n)=(2b_1+1)(2b_2+1)...(2b_i+1+1)...(2b_k+1)$.
$Rightarrow 2|tau (n)$
$color{red}{textrm{However we know that every perfect square has an odd number of factors.}}$ This contradicts our assumption that $n$ is a perfect square.
Thus, our assumption is wrong and no such $a_i=2b_i+1$ exists.
QED.
Can the assumption highlighted in red be generalized for any number which is an $m^{th}$ power? And if not, is there a better way of doing it than my proof?
elementary-number-theory proof-verification
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
$endgroup$
2
$begingroup$
That is very roundabout. How about simply: "Suppose $n=b^m$ for some $b$. Then $b$ has a prime factorization ... (bla bla) ... and therefore $n$ has the following prime factorization ... (bla bla) ... and since prime factorizations are unique ..."
$endgroup$
– Henning Makholm
Jan 12 at 6:02
$begingroup$
I tried it that way, but that leads me to the statement $n=(q_1q_2...q_k)^mp_i^{a_i}$ where $m$ does not divide $a_i$. Can I just dismiss this as not being an $m^{th}$ power? I assumed that there was no way to say that.
$endgroup$
– Naman Kumar
Jan 12 at 6:15
$begingroup$
How on earth do you get that statement? What is the prime factorization you assume for $b$? Why are you treating a particular $p_i$ specially?
$endgroup$
– Henning Makholm
Jan 12 at 6:19
$begingroup$
I'm sorry, I misunderstood what you were saying. Can you elaborate on that statement 'by the uniqueness of prime factorisation...'?
$endgroup$
– Naman Kumar
Jan 12 at 6:23
2
$begingroup$
If you have computed one prime factorization of $b^m$ where all of the exponents are multiples of $m$, then you know that this is the prime factorization of $n$ (because prime factorizations are unique), and therefore the $alpha_i$s must be the numbers you have just concluded are multiples of $m$.
$endgroup$
– Henning Makholm
Jan 12 at 6:26
|
show 1 more comment
$begingroup$
I am a high school student self-studying number theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics:
Prove that any integer $n$ whose prime factorization is $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ is a perfect square iff $2|a_i$, and in general $n$ is an $m^{th}$ power iff all $a_i$ is a multiple of $m$.
I managed to prove this for $m=2$, as the question said, but am unable to even get started on the second part, because my proof required an assumption which I do not know if holds for the general case also.
Please check the proof below:
$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ where $n$ is a perfect square.
Let us assume, if possble, that there exist some powers $a_igeq 0$ such that $a_i=2b_i+1$.
We know that the number of factors $tau(n)=(a_1+1)(a_2+1)...(a_k+1)$. We can write each $a_i$ as $a_i= 2b_i$ or $a_i=2b_i+1$.
Now $tau(n)=(2b_1+1)(2b_2+1)...(2b_i+1+1)...(2b_k+1)$.
$Rightarrow 2|tau (n)$
$color{red}{textrm{However we know that every perfect square has an odd number of factors.}}$ This contradicts our assumption that $n$ is a perfect square.
Thus, our assumption is wrong and no such $a_i=2b_i+1$ exists.
QED.
Can the assumption highlighted in red be generalized for any number which is an $m^{th}$ power? And if not, is there a better way of doing it than my proof?
elementary-number-theory proof-verification
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
$endgroup$
I am a high school student self-studying number theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics:
Prove that any integer $n$ whose prime factorization is $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ is a perfect square iff $2|a_i$, and in general $n$ is an $m^{th}$ power iff all $a_i$ is a multiple of $m$.
I managed to prove this for $m=2$, as the question said, but am unable to even get started on the second part, because my proof required an assumption which I do not know if holds for the general case also.
Please check the proof below:
$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ where $n$ is a perfect square.
Let us assume, if possble, that there exist some powers $a_igeq 0$ such that $a_i=2b_i+1$.
We know that the number of factors $tau(n)=(a_1+1)(a_2+1)...(a_k+1)$. We can write each $a_i$ as $a_i= 2b_i$ or $a_i=2b_i+1$.
Now $tau(n)=(2b_1+1)(2b_2+1)...(2b_i+1+1)...(2b_k+1)$.
$Rightarrow 2|tau (n)$
$color{red}{textrm{However we know that every perfect square has an odd number of factors.}}$ This contradicts our assumption that $n$ is a perfect square.
Thus, our assumption is wrong and no such $a_i=2b_i+1$ exists.
QED.
Can the assumption highlighted in red be generalized for any number which is an $m^{th}$ power? And if not, is there a better way of doing it than my proof?
elementary-number-theory proof-verification
elementary-number-theory proof-verification
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
edited Jan 12 at 6:01
Naman Kumar
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
asked Jan 12 at 5:54
Naman KumarNaman Kumar
22813
22813
2
$begingroup$
That is very roundabout. How about simply: "Suppose $n=b^m$ for some $b$. Then $b$ has a prime factorization ... (bla bla) ... and therefore $n$ has the following prime factorization ... (bla bla) ... and since prime factorizations are unique ..."
$endgroup$
– Henning Makholm
Jan 12 at 6:02
$begingroup$
I tried it that way, but that leads me to the statement $n=(q_1q_2...q_k)^mp_i^{a_i}$ where $m$ does not divide $a_i$. Can I just dismiss this as not being an $m^{th}$ power? I assumed that there was no way to say that.
$endgroup$
– Naman Kumar
Jan 12 at 6:15
$begingroup$
How on earth do you get that statement? What is the prime factorization you assume for $b$? Why are you treating a particular $p_i$ specially?
$endgroup$
– Henning Makholm
Jan 12 at 6:19
$begingroup$
I'm sorry, I misunderstood what you were saying. Can you elaborate on that statement 'by the uniqueness of prime factorisation...'?
$endgroup$
– Naman Kumar
Jan 12 at 6:23
2
$begingroup$
If you have computed one prime factorization of $b^m$ where all of the exponents are multiples of $m$, then you know that this is the prime factorization of $n$ (because prime factorizations are unique), and therefore the $alpha_i$s must be the numbers you have just concluded are multiples of $m$.
$endgroup$
– Henning Makholm
Jan 12 at 6:26
|
show 1 more comment
2
$begingroup$
That is very roundabout. How about simply: "Suppose $n=b^m$ for some $b$. Then $b$ has a prime factorization ... (bla bla) ... and therefore $n$ has the following prime factorization ... (bla bla) ... and since prime factorizations are unique ..."
$endgroup$
– Henning Makholm
Jan 12 at 6:02
$begingroup$
I tried it that way, but that leads me to the statement $n=(q_1q_2...q_k)^mp_i^{a_i}$ where $m$ does not divide $a_i$. Can I just dismiss this as not being an $m^{th}$ power? I assumed that there was no way to say that.
$endgroup$
– Naman Kumar
Jan 12 at 6:15
$begingroup$
How on earth do you get that statement? What is the prime factorization you assume for $b$? Why are you treating a particular $p_i$ specially?
$endgroup$
– Henning Makholm
Jan 12 at 6:19
$begingroup$
I'm sorry, I misunderstood what you were saying. Can you elaborate on that statement 'by the uniqueness of prime factorisation...'?
$endgroup$
– Naman Kumar
Jan 12 at 6:23
2
$begingroup$
If you have computed one prime factorization of $b^m$ where all of the exponents are multiples of $m$, then you know that this is the prime factorization of $n$ (because prime factorizations are unique), and therefore the $alpha_i$s must be the numbers you have just concluded are multiples of $m$.
$endgroup$
– Henning Makholm
Jan 12 at 6:26
2
2
$begingroup$
That is very roundabout. How about simply: "Suppose $n=b^m$ for some $b$. Then $b$ has a prime factorization ... (bla bla) ... and therefore $n$ has the following prime factorization ... (bla bla) ... and since prime factorizations are unique ..."
$endgroup$
– Henning Makholm
Jan 12 at 6:02
$begingroup$
That is very roundabout. How about simply: "Suppose $n=b^m$ for some $b$. Then $b$ has a prime factorization ... (bla bla) ... and therefore $n$ has the following prime factorization ... (bla bla) ... and since prime factorizations are unique ..."
$endgroup$
– Henning Makholm
Jan 12 at 6:02
$begingroup$
I tried it that way, but that leads me to the statement $n=(q_1q_2...q_k)^mp_i^{a_i}$ where $m$ does not divide $a_i$. Can I just dismiss this as not being an $m^{th}$ power? I assumed that there was no way to say that.
$endgroup$
– Naman Kumar
Jan 12 at 6:15
$begingroup$
I tried it that way, but that leads me to the statement $n=(q_1q_2...q_k)^mp_i^{a_i}$ where $m$ does not divide $a_i$. Can I just dismiss this as not being an $m^{th}$ power? I assumed that there was no way to say that.
$endgroup$
– Naman Kumar
Jan 12 at 6:15
$begingroup$
How on earth do you get that statement? What is the prime factorization you assume for $b$? Why are you treating a particular $p_i$ specially?
$endgroup$
– Henning Makholm
Jan 12 at 6:19
$begingroup$
How on earth do you get that statement? What is the prime factorization you assume for $b$? Why are you treating a particular $p_i$ specially?
$endgroup$
– Henning Makholm
Jan 12 at 6:19
$begingroup$
I'm sorry, I misunderstood what you were saying. Can you elaborate on that statement 'by the uniqueness of prime factorisation...'?
$endgroup$
– Naman Kumar
Jan 12 at 6:23
$begingroup$
I'm sorry, I misunderstood what you were saying. Can you elaborate on that statement 'by the uniqueness of prime factorisation...'?
$endgroup$
– Naman Kumar
Jan 12 at 6:23
2
2
$begingroup$
If you have computed one prime factorization of $b^m$ where all of the exponents are multiples of $m$, then you know that this is the prime factorization of $n$ (because prime factorizations are unique), and therefore the $alpha_i$s must be the numbers you have just concluded are multiples of $m$.
$endgroup$
– Henning Makholm
Jan 12 at 6:26
$begingroup$
If you have computed one prime factorization of $b^m$ where all of the exponents are multiples of $m$, then you know that this is the prime factorization of $n$ (because prime factorizations are unique), and therefore the $alpha_i$s must be the numbers you have just concluded are multiples of $m$.
$endgroup$
– Henning Makholm
Jan 12 at 6:26
|
show 1 more comment
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2
$begingroup$
That is very roundabout. How about simply: "Suppose $n=b^m$ for some $b$. Then $b$ has a prime factorization ... (bla bla) ... and therefore $n$ has the following prime factorization ... (bla bla) ... and since prime factorizations are unique ..."
$endgroup$
– Henning Makholm
Jan 12 at 6:02
$begingroup$
I tried it that way, but that leads me to the statement $n=(q_1q_2...q_k)^mp_i^{a_i}$ where $m$ does not divide $a_i$. Can I just dismiss this as not being an $m^{th}$ power? I assumed that there was no way to say that.
$endgroup$
– Naman Kumar
Jan 12 at 6:15
$begingroup$
How on earth do you get that statement? What is the prime factorization you assume for $b$? Why are you treating a particular $p_i$ specially?
$endgroup$
– Henning Makholm
Jan 12 at 6:19
$begingroup$
I'm sorry, I misunderstood what you were saying. Can you elaborate on that statement 'by the uniqueness of prime factorisation...'?
$endgroup$
– Naman Kumar
Jan 12 at 6:23
2
$begingroup$
If you have computed one prime factorization of $b^m$ where all of the exponents are multiples of $m$, then you know that this is the prime factorization of $n$ (because prime factorizations are unique), and therefore the $alpha_i$s must be the numbers you have just concluded are multiples of $m$.
$endgroup$
– Henning Makholm
Jan 12 at 6:26